How do you find the polar equation for x^2+(y-2)^2-4=0#?

Question

Charles Autterson · Accepted Answer

#r=4sintheta# The relation between polar coordinates #(r,theta)# and rectangular coordinates #(x,y)# are given by #x=rcosthata# and #y=rsintheta# and hence #r=sqrt(x^2+y^2)# Hence #x^2+(y-2)^2-4=0# can be written as #x^2+y^2-4y+4-4=0# or using relations between polar and Cartesian coordinates #x^2+y^2-4y=0# #r^2-4rsintheta=0# or #r-4sintheta=0# or #r=4sintheta#

James Adkins · Answer

undefined To find the polar equation for $x^2 + (y - 2)^2 - 4 = 0$, you can substitute $x = r\cos(\theta)$ and $y = r\sin(\theta)$ into the equation, where $r$ represents the radial distance from the origin and $\theta$ is the angle measured counterclockwise from the positive x-axis.

Substituting these expressions for $x$ and $y$ into the given equation yields:

$$
(r\cos(\theta))^2 + ((r\sin(\theta)) - 2)^2 - 4 = 0
$$

Expand and simplify this equation:

$$
r^2\cos^2(\theta) + (r\sin(\theta) - 2)^2 - 4 = 0
$$

$$
r^2\cos^2(\theta) + r^2\sin^2(\theta) - 4r\sin(\theta) + 4 - 4 = 0
$$

Using the trigonometric identity $ \cos^2(\theta) + \sin^2(\theta) = 1 $, we have:

$$
r^2 - 4r\sin(\theta) = 0
$$

Now, we solve for $r$:

$$
r(r - 4\sin(\theta)) = 0
$$

This equation represents two curves: $r = 0$ (the pole) and $r = 4\sin(\theta)$. However, $r = 0$ corresponds to the origin, and $r = 4\sin(\theta)$ is the polar equation for the given Cartesian equation.

Thus, the polar equation for $x^2 + (y - 2)^2 - 4 = 0$ is $ r = 4\sin(\theta) $.